Question

A particle moving witj constant acceleration 2 m/s² due west has an initial velocity 9 m/s due west .Find distance covered in 5th second of its motion is​

Answer 1

EXPLANATION.

Particle moving with constant acceleration = 2 m/s² due to west.

initial velocity = 9 m/s due to west.

As we know that,

Formula of :

Newton first equation of motion.

⇒ v = u + at.

Newton second equation of motion.

⇒ S = ut + 1/2at².

S(nth) = u + a/2(2n - 1).

Using this formula in the equation, we get.

To find : 5th seconds.

⇒ S(5th) = 9 + (-2)/2)[2(5) - 1].

⇒ S(5th) = 9 + (-1)(9).

⇒ S(5th) = 9 - 9.

⇒ S(5th) = 0.

Displacement on 5th seconds = 0.

Initial velocity = u = 9 m/s.

Final velocity = v = 0.

acceleration = a = 2 m/s².

⇒ v = u + at.

⇒ 0 = 9 - 2(t).

⇒ 9 = 2t.

⇒ t = 4.5 seconds.

Now, we find velocity at 4 seconds.

⇒ v = u + at.

⇒ v = 9 - 2 x 4.

⇒ v = 9 - 8.

⇒ v = 1 m/s.

Now, we find distance from 4 seconds to 4.5 seconds.

⇒ Δt = 4.5 - 4 = 0.5 seconds.

⇒ S = ut + 1/2at².

⇒ S = (1)(0.5) - 1/2x (2) x (0.5)².

⇒ S = 0.5 - 0.25.

⇒ S = 0.25 m.

Now, we find distance from 4.5 seconds to 5 seconds.

⇒ Δt = 5 - 4.5 = 0.5 seconds.

Velocity will be zero from 4.5 seconds to 5 seconds.

⇒ S = ut + 1/2at².

⇒ S = 0 + 1/2 x 2 x (0.5)².

⇒ S = 0.25 m.

To find :

Distance covered in 5th seconds.

Distance covered in 4 to 4.5 seconds + Distance covered in 4.5 to 5 seconds.

⇒ 0.25 m + 0.25 m = 0.50 m.

Distance covered in 5th seconds = 0.50 m.

Answer 2

Answer:

Acceleration = 2m/s² in West direction

In vector form ,

$\rm :\dashrightarrow{a = -2i \: m/s²}$

Initial velocity = 9 m/s² in East direction ,

In vector form ,

$\rm :\dashrightarrow{u = 9i \: m/s}$

Use kinematic equation ,

$\boxed{\overbrace{\tt \red{Snth = u + \dfrac{a(2n - 1)}{2}} }}$

Here Snth , u , and a all are in vector form

Snth is displacement in nth second

u is intial velocity of particle

a is acceleration of particle .

$\sf{Snth = 9i + \dfrac{(-2i)(2 × 5 - 1)}{2}}$

$\sf :\longmapsto{ 9i + \dfrac{(-2i)(9)}{2}}$

$\sf :\longmapsto{9i - \dfrac{18i}{2}}$

$\sf :\longmapsto{9i - 9i }$

$\bold{\underline{\sf \purple{ :\dashrightarrow{ 0}}}}$

Hence, particle displacement in 5th second = 0 .

particle displacement in 5th second = 0 .it means particle reached 5th second in intial position .

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For distance :-

Here we see displacement = 0 in 5th second . it means in 5th second final position = initial position.

First we find out where ,

velocity becomes zero

use ,

$\tt{V = u + at }$

$\tt{0 = 9 -2t }$

$\tt{t = 4.5 \: sec }$

Now ,

Find out velocity at 4 sec

$\tt{v = u + at }$

$\tt{v = 9 - 2 × 4 = 1m/s}$

So, distance covered in 4sec to 4.5 sec

$\sf :\longmapsto{ut + \frac{1}{2}at² }$

$\sf :\longmapsto{ 1 × \frac{1}{2} - \frac{1}{2} × 2 × \frac{1}{4} }$

$\sf :\longmapsto{0.25 m}$

Again,

Distance covered in 4.5 sec to 5 sec ,

velocity at 4.5 sec = 0

So,

Distance = $\rm{\dfrac{1}{2}at^{2}}$

$\sf :\longmapsto{\frac{1}{2} × 2 × \frac{1}{4}}$

$\bf\color{magenta}{ :\longmapsto {0.25 m}}$

Hence, distance covered in 5th second = Distance covered in 4 sec to 4.5 sec + distance Covered in 4.5 sec to 5sec

$\qquad \qquad \sf :\longmapsto{ 0.25m + 0.25m}$

$\qquad \qquad \bf\color{magenta}{:\longmapsto{0.5 m}}$

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HOPE IT HELPS!:)